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Shtirlitz [24]
3 years ago
14

Convert 330 degrees to radians. Plz help

Mathematics
1 answer:
geniusboy [140]3 years ago
7 0
5.7595. i am not sure if i’m correct but it is the rounded version.
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Answer:

X=3

Step-by-step explanation:

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PLSSS HELP IF YOU TURLY KNOW THISS
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I think the answer is 8.25 × 10^3

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5(−6)= ________ <br> What is the answer?
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I need help with 1-3 please and help!
Ghella [55]

Answer:

Q : 3

10x - 11 = 120 - 11 = 109°

 3x - 2    = 36 - 2  = 34°

 3x + 1     = 36 + 1  = 37°

Q ; 2

3x - 5 = 27 - 5= 22°

7x + 5 = 63 + 5 = 68°

And 90°

Q1 :

∠1 = 92°

∠2 = 42°

∠3 =  113°

Step-by-step explanation:

Solution for Q : 3

As the angle of all three is given as ,

10x - 11

3x - 2

3x + 1

We know sum of all the three angles of triangle = 180 °

So, (10x - 11) + (3x - 2) + (3x + 1) = 180°

Or,   16x - 12 = 180°

Or     16x = 192°, So    , x = 12

So, all three angles are 10x - 11 = 120 - 11 = 109°

                                       3x - 2    = 36 - 2  = 34°

                                       3x + 1     = 36 + 1  = 37°

Solution for Q - 2

Given angles are

3x - 5

7x + 5

90°

We know sum of all the three angles of triangle = 180 °

so ,(3x - 5) + (7x + 5) + 90 = 180°

or 10x  =                       180 - 90 = 90°

SO, x = 9°

SO, all the three angles are 3x - 5 = 27 - 5= 22°

                                              7x + 5 = 63 + 5 = 68°

And                                                                    90°

Solution for Q : 1

From,

the shown fig it is clear that

The ∠2 = 42°      (<u> opposite vertical angles</u> )

so, in  left triangle

50° + ∠2 + ∠1  = 180°

Or,  50° + 42° + ∠1 = 180°     ( sum of all angles of triangles = 180°)

Or, ∠1 = 92°

 Again

From right figure triangle

∠2 + 25° + ∠3 = 180°

Or, 42° + 25° + ∠3 = 180

Or, ∠3 = 113°

3 0
3 years ago
Prove that in the plane the perpendicular bisector of a chord in a circle contains the center of the circle
katrin [286]

Answer:

The cord of a circle is a segment whose endpoints are on the circle. This means that the perpendicular bisector of any chord passes through the center of the circle.

3 0
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